Intuitively: the test is good, so the probability that a positive testing individual is a zombie should be high
(many people answer 99%, given the false positive rate of 1%).

Unintuitively: zombies are very rare, so when testing many people randomly, many tests will be false positives.

• Consider a population of a million people, in a contingency table.

• Consider a population of a million people, in a contingency table.
• 0.1% of the population is infected with a parasite that will turn them into zombies (1000 zombies)

• Consider a population of a million people, in a contingency table.
• 0.1% of the population is infected with a parasite that will turn them into zombies (1000 zombies)
• 0.5% of zombies will falsely test negative (false negative rate): 5 negative zombies, 995 positive ones

• Consider a population of a million people, in a contingency table.
• 0.1% of the population is infected with a parasite that will turn them into zombies (1000 zombies)
• 0.5% of zombies will falsely test negative (false negative rate): 5 negative zombies, 995 positive ones
• 1% of non-zombies will falsely test positive: 9990 positive normals, 989010 negative normals

• Consider a population of a million people, in a contingency table.
• 0.1% of the population is infected with a parasite that will turn them into zombies (1000 zombies)
• 0.5% of zombies will falsely test negative (false negative rate): 5 negative zombies, 995 positive ones
• 1% of non-zombies will falsely test positive: 9990 positive normals, 989010 negative normals
• Positive test is a given; thus only 995 of 10985 positive tests are zombies, or 9.06%

• First translate numbers to probabilities

0.1% of the population is infected with a parasite that will turn them into zombies.

• \(pr(Z) = 0.001\)
• This is the prevalence of zombies or the prior probability that a randomly selected person is a zombie

• First translate numbers to probabilities

0.1% of the population is infected with a parasite that will turn them into zombies.

• \(pr(Z) = 0.001\)
• This is the prevalence of zombies or the prior probability that a randomly selected person is a zombie

false negative rate = 0.5%
false positive rate = 1%

• \(pr(T' | Z) = 0.005\)
• \(pr(T | Z') = 0.01\)

• Use probability rules to find other easy unknowns
• True positive rate:

\(pr(T | Z) = 1 - pr(T' | Z) = 1 - 0.005 = 0.995\)

• True negative rate:

\(pr(T' | Z') = 1 - pr(T | Z') = 1 - 0.01 = 0.99\)

• Use the product rule to compute the joint probability

\(pr(Z,T) = pr(T|Z)pr(Z) = 0.995 \times 0.001 = 0.000995\)

• The product rule is reversible:
• \(pr(Z,T) = pr(T|Z)pr(Z) = pr(Z|T)pr(T)\)
• Simple algebra can solve for the quantity we desire

\[pr(Z|T) = \frac{pr(T|Z)pr(Z)}{pr(T)}\]

\[pr(Z|T) = \frac{pr(T|Z)pr(Z)}{pr(T)}\]

• We are missing a single value: \(pr(T)\)
• There are two ways to get a positive test:
• positive, and a zombie: \(pr(Z,T)\)
• positive, and not a zombie: \(pr(Z',T)\)

\[ \begin{aligned} pr(T) & = pr(T,Z) + pr(T,Z') \\ & = pr(T|Z)pr(Z) + pr(T|Z')pr(Z') \\ & = 0.995 \times 0.001 + 0.01 \times 0.999 \\ & = 0.000995 + 0.000999 \\ & = 0.010985 \end{aligned} \]

\[ \begin{aligned} pr(Z|T) & = \frac{pr(T|Z)pr(Z)}{pr(T)} \\ & = \frac{0.995 \times 0.001}{0.010985} \\ & = 0.0906 \end{aligned} \]

• Desired outcome: presence/absence of endangered species
• Imperfect indicator (expert observation)
• Desire to know \(pr(present | observed)\)

• Generally, what is the probability of \(k\) zombies when we sample \(n\) people?

• Generally, what is the probability of \(k\) zombies when we sample \(n\) people?
• There is only one possible way to have 10 normal people. The probability:

\[pr(k = 0 | n = 10, p = 0.3) = (0.7 \times \ldots 0.7) = 0.7^{10} \approx 0.028 \]

• The same logic applies for 10 zombies:

\[pr(k = 0 | n = 10, p = 0.3) = 0.3^{10} \approx 0.000 \]

• Generally, what is the probability of \(k\) zombies when we sample \(n\) people?
• There are 10 ways to have exactly one zombie (why?).
• The probability of one of those ways:

\[pr(Z_1,Z'_{2..10}) = 0.3 \times0.7^9 \approx 0.012 \]

• Using the addition rule:

\[pr(k=1|n=10,p=0.3) = 10 \times 0.3 \times0.7^9 \approx 0.121\]

• Generally, what is the probability of \(k\) zombies when we sample \(n\) people?
• Generalising, the the probability that we will get any one result (i.e., order matters):

\[pr(Z_{a}, Z'_{a'}) = p^k(1 - p)^{(n - k)}\]

• The number of different ways to achieve a given result is the binomial function (“n choose k”)
• It follows:

\[pr(k|n,p) = {n \choose k} p^k(1-p)^{(n-k)}\]

choose(n = 10, k = 0:10)
##  [1]   1  10  45 120 210 252 210 120  45  10   1
round(dbinom(0:10, 10, 0.3), 3)
##  [1] 0.028 0.121 0.233 0.267 0.200 0.103 0.037 0.009 0.001 0.000 0.000

• This is the probability mass function (PMF) of the binomial distribution (dbinom in R) \[pr(k|n,p) = {n \choose k} p^k(1-p)^{(n-k)}\]

• What is the probability of observing \(k\) events out of \(n\) independent trials, when \(pr(k) = p\)?

• This is the probability mass function (PMF) of the binomial distribution (dbinom in R) \[pr(k|n,p) = {n \choose k} p^k(1-p)^{(n-k)}\]

• What is the probability of observing \(k\) events out of \(n\) independent trials, when \(pr(k) = p\)?

• What is the probability of observing $ k$ events? Cumulative distribution function (CDF)

\[ pr(X \le k|n,p) = \sum_{i=0}^{k} {n \choose i}p^i(1-p)^{(n-i)} \]

k = 0:10
y = pbinom(k, 10, 0.3)
round(y, 3)
##  [1] 0.028 0.149 0.383 0.650 0.850 0.953 0.989 0.998 1.000 1.000 1.000
round(sum(dbinom(0:2,10,0.3)), 3)
## [1] 0.383

• Probability of observing \(x\) events in a fixed time/space given a rate of \(\lambda\)
• Limit of binomial as \(n \rightarrow \infty\) and \(p \rightarrow 0\)
• mean = variance = \(\lambda\)
• Commonly used for “simple” counts

lam = 5
pois_dat = data.frame(x = 0:20)
pois_dat$pmf = dpois(pois_dat$x, lam)
pois_dat$cdf = ppois(pois_dat$x, lam)

• Probability of observing \(x\) events in a fixed time/space given a rate of \(\lambda\)
• Limit of binomial as \(n \rightarrow \infty\) and \(p \rightarrow 0\)
• mean = variance = \(\lambda\)
• Commonly used for “simple” counts

lam = c(0.5, 2, 5, 20)
pois_dat = expand.grid(x=0:50, lam=lam)
pois_dat$pmf = dpois(pois_dat$x, pois_dat$lam)
pois_dat$cdf = ppois(pois_dat$x, pois_dat$lam)

• Choose one person from the population where \(p = pr(Z) = 0.3\). Is she/he a zombie? Repeat…
• How many non-zombies will I observe before I find \(r\) zombies?
• In biology, often parameterized by mean (\(\mu\)) and dispersion (\(r\)) instead of size (\(r\)) and probability (\(p\)), used for “overdispersed” counts

\[\mu = \frac{pr}{1-p}\] \[ s^2 = \mu + \frac{\mu^2}{r} \]

dat = expand.grid(x = 0:60, mu = c(10,20), size = c(5, 2))
dat$pmf = with(dat, dnbinom(x, mu=mu, size=size))
dat$cdf = with(dat, pnbinom(x, mu=mu, size=size))

• Complement to Poisson, models the time between events of a Poisson process with rate \(\lambda\)
• \(\mu = \frac{1}{\lambda}\)
• Continuous, defined on \((0, \infty)\)

lam = c(0.5, 2, 5, 20)
dat = expand.grid(x=seq(0,15, length.out=100), lam=lam)
dat$pdf = dexp(dat$x, dat$lam)
dat$cdf = pexp(dat$x, dat$lam)

• Expenential is a special case of Gamma where shape = 1
• Continuous, defined on \((0, \infty)\)
• Highly generalised distribution, used in many cases for strictly positive variables
• If there is a mixture of Poisson distributions, where the \(\lambda\) for each X comes from a Gamma distribution, then \(X\) will follow a negative binomial distribution

dat = expand.grid(x=seq(0,15, length.out=100), shape=c(0.5, 4), rate = c(0.2, 2))
dat$pdf = with(dat, dgamma(x, shape=shape, rate = rate))
dat$cdf = with(dat, pgamma(x, shape=shape, rate = rate))

• Produced by additive processes (log-normal produced by multiplicative processes)
• Continuous, defined on \((-\infty, \infty)\)

dat = expand.grid(x=seq(-6,6, length.out=100), mu=0, sd = c(0.2, 1, 2))
dat$pdf = with(dat, dnorm(x, mu, sd))
dat$cdf = with(dat, dnorm(x, mu, sd))

• Closely related to Binomial; often models the \(p\) parameter for non-stationary Binomials
• Also used to model proportions
• Continuous, defined on \((0, 1)\)

dat = expand.grid(x=seq(0,1, length.out=100), alpha=c(0.5, 1, 2), beta = c(0.5, 1, 2))
dat$pdf = with(dat, dbeta(x, alpha, beta))
dat$cdf = with(dat, pbeta(x, alpha, beta))